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Fofino [41]
2 years ago
13

For the reaction 2NH3(g) + O2(g) -> 2NO(g) + 3H2O(g). The reaction is run using 824 g NH3 and excess oxygen. How many moles o

f NO are formed? How many moles of H2O are formed?
Chemistry
1 answer:
Alex73 [517]2 years ago
4 0

Answer:

Explanation:

Step One

Find the molar Mass of NH3

N =   14

3H =  3

Mass = 17

Step Two

Find the number of moles of NH3

mols = given mass / Molar Mass

given mass = 824 grams

molar mass = 17

mols = 824/17

mols =  48.47

Step Three

Find the moles of NO

Here you have to look at the balanced equation

Every 2 mols of HN3 produces 2 mols of NO

So the number of mols of NH3 = 48.47

The number of mols of  NO      = 48.47

Step Four

Find the mols of H2O

2 mols of NH3 produces 3 moles of H20

48.47 mols H20 produces x moles of H20

3/x = 2/48.47     Cross Multiply

2x = 3*48.47

2x = 145.41         Divide by 2

x = 145.41/2

x = 72.71            mols of water.

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Answer:

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Explanation:

An oxidizing agent is an element in a reaction that accepts the electrons of another element. It is typically hydrogen, oxide, or any halogen. In this case, it is oxygen. The answer is 02.

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You have three elements, A, B, and C, with the following electronegativity values:
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#AB

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#AC

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6 0
2 years ago
What is the pH of a solution made by mixing 200 mL of 0.025M HCl and 150 mL of 0.050 M HCl?
seraphim [82]

Answer:

The answer to your question is pH = 1.45  

Explanation:

Data

pH = ?

Volume 1 = 200 ml

[HCl] 1 = 0.025 M

Volume 2 = 150 ml

[HCl] 2 = 0.050 M

Process

1.- Calculate the number of moles of each solution

Solution 1

                Molarity = moles / volume

-Solve for moles

                moles = 0.025 x 0.2

result

               moles = 0.005

Solution 2

               moles = 0.050 x 0.15

-result

                moles = 0.0075

2.- Sum up the number of moles

Total moles = 0.005 + 0.0075

                   = 0.0125

3.- Sum up the volume

total volume = 200 + 150

                     350 ml or 0.35 l

4.- Calculate the final concentration

Molarity = 0.0125 / 0.35

              = 0.0357

5.- Calculate the pH

pH = -log [H⁺]

-Substitution

pH = -log[0.0357]

-Result

pH = 1.45    

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